Question

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular acceleration of 0.5 rad.s?. The magnitude of the applied torque is: a) 0.30 N.m b) 3.0 x 102 N.m C) 0.15 N.m d) 7.5 x 102 N.m e) 6.0 x 102 N.m.

Answer 1

Answer:

The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

$\tau=I\times\alpha$

Here, I = mk²

$\tau=mk^2\times\alpha$

Put the value into the formula

$\tau=3\times(0.2)^2\times0.5$

$\tau=0.06\ N-m$

$\tau=6.0\times10^{-2}\ N-m$

Hence, The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$